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Figure shows the circuit of a potentiome...

Figure shows the circuit of a potentiometer. The length of the potentiometer wire AB is 50 cm. The emf of the battery E, is 4 volt having negligible internal resistance. Values of resistances `R_(1)` and `R_(2)` are `15` ohm and `5` ohm respectively. When both the keys are open, he null point is obtained at a distance of `31.25` cm from end A but when both the keys are closed, the balance length reduces to 5 cm only `R_(AB) = 10 Omega `

The emf of the cell `E_(2)` is

A

1 volt

B

2 volt

C

3 volt

D

4 volt

Text Solution

Verified by Experts

The correct Answer is:
A
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Figure shows the circuit of a potentiometer. The length of the potentiometer wire AB is 50 cm. The emf of the battery E, is 4 volt having negligible internal resistance. Values of resistances R_(1) and R_(2) are 15 ohm and 5 ohm respectively. When both the keys are open, he null point is obtained at a distance of 31.25 cm from end A but when both the keys are closed, the balance length reduces to 5 cm only R_(AB) = 10 Omega The internal resistance of cell E_(2) is

Figure shows the circuit of a potentiometer. The length of the potentiometer wire AB is 50 cm. The emf of the battery E, is 4 volt having negligible internal resistance. Values of resistances R_(1) and R_(2) are 15 ohm and 5 ohm respectively. When both the keys are open, he null point is obtained at a distance of 31.25 cm from end A but when both the keys are closed, the balance length reduces to 5 cm only R_(AB) = 10 Omega Which of the following can be possible way to connect the batteries in the potentiometer setup above ?

Knowledge Check

  • A potentiometer wire of length 100 cm has a resistance of 100Omega it is connected in series with a resistance and a battery of emf 2 V and of negligible internal resistance. A source of emf 10 mV is balanced against a length of 40 cm of the potentiometer wire. what is the value of the external resistance?

    A
    `790Omega`
    B
    `890Omega`
    C
    `990Omega`
    D
    `1090Omega`

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